# Poles and zeros of meromorphic function. problem from Conway's book

Let $$f$$ be meromorphic on $$G$$. Show that neither the poles nor the zeros of $$f$$ have a limit point in $$G$$.

Proof: 1. Let $$Z=\{\text{zeros of } f\}$$ and suppose that $$Z$$ has a limit point $$b$$ in $$G$$. Then exists $$\{z_n\}\in Z$$, $$z_n$$ are distinct and $$z_n\neq b$$ such that $$z_n\to b$$. Then $$f(z_n)\to f(b)$$ and hence $$f(b)=0$$. But we know that $$f\equiv 0$$ iff $$\{z: f(z)=0\}$$ has a limit point in $$G$$. Hence we get that $$f\equiv 0$$ but this is not the case since $$f$$ is meromorphic.

1. Let $$P=\{\text{poles of } f\}$$ and suppose that $$P$$ has a limit point $$a$$ in $$G$$. Then exists $$\{p_n\}\in P$$, $$p_n$$ are distinct and $$p_n\neq a$$ such that $$p_n\to a$$. Since meromorphic function we can consider as continuous in $$\mathbb{C}\cup \{\infty\}$$ then $$f(p_n)\to f(a)$$ and hence $$|f(a)|=\infty$$. So $$a$$ is a pole of $$f(z)$$. Hence $$f(z)(z-a)^k=g(z)$$ for some $$k\geq 1$$ and $$g$$ is analytic function. If we plug in any point $$p_n$$ we get that $$g(p_n)=\infty$$ which is not the case.

Is my solution correct?

Would be very grateful for remarks!

• Not correct. Nothing prevents $g$ from having a pole at $p_n$. – user647486 Apr 14 '19 at 17:00
• Analytic where? You haven't said. Certainly not on $G$. For example, if you take $f(z)=\frac{1}{(z-1)(z-2)}$ and $a=2$. Then $f(z)(z-2)^1=\frac{1}{z-1}=g(z)$ still has a pole at $z=1$. – user647486 Apr 14 '19 at 17:12
• The proofs (1) and (2) are equivalent by working with $f$ and $1/f$ respectively and applying the identity principle. You can just deduce (2) by applying (1) to $1/f$. – user647486 Apr 14 '19 at 17:18
• By the way, I hadn't noticed. The statement in the first sentence is not correct. Probably not properly quoted. The identity functions $f\equiv0$ and $f\equiv\infty$ are both meromorphic. Unless for some reason Conway deliberately excluded constant functions from his definition of meromorphic, which is not a common practice. – user647486 Apr 14 '19 at 17:23
• Well, I can't. The statement is false as written. I checked his book and you didn't misquote him and his definition doesn't exclude constant functions. The statement should've been "Let $f$ be a non-constant meromorphic function on $G$ ...". For that statement, your proof of (1) is correct. – user647486 Apr 14 '19 at 17:28

For the zeros you have a minor gap.

A meromorphic function $$f$$ on $$G$$ is a function that is holomorphic on all of $$G$$ except for a set $$P$$ of isolated points which are poles of $$f$$. The set $$P$$ is closed in $$G$$, hence $$G' = G \setminus P$$ is open in $$G$$ and therefore open in $$\mathbb C$$. Now $$f$$ is defined on all of $$G'$$. If the limit point $$b$$ of $$Z$$ is contained in $$G'$$, your agument works. So you have to show that $$b \notin P$$. But if $$b \in P$$, then $$\lvert f(z)\rvert > 0$$ for $$z \in G$$ with $$\lvert z - b \rvert < \epsilon$$ which is impossible since $$\lvert z_n - b \rvert < \epsilon$$ for $$n \ge n_0$$.

To deal with poles you can argue as follows. The function $$g = \frac{1}{f}$$ is defined on $$G' \setminus Z$$. From the first step we know that $$Z$$ is a set of isolated points. Moreover, since $$f \ne 0$$, all zeros must have finite order and we conclude that $$g$$ has a pole at each $$z \in Z$$. The singularities of $$g$$ at the points $$p \in P$$ are removable because $$\lim_{z \to p} \lvert g(z) \rvert = \lim_{z \to p} \frac{1}{\lvert f(z) \rvert} = 0$$. Hence $$g$$ extends to a homolomorphic function on $$G \setminus Z$$, in other words a meromorphic function on $$G$$ with zeros at all $$p \in P$$. Now apply again the first step.

The general picture in step 2 is this.

Let $$\mathfrak{S}(G)$$ denote the set of all pairs $$(S,f)$$ such that $$S$$ is a set of isolated points in $$G$$ and $$f : G \setminus S \to \mathbb C$$ is holomorphic function having either a removable singularity or a pole at each $$s \in S$$. Let $$R(f)$$, $$P(f)$$ and $$Z(f)$$ denote the set of removable singularities, poles and zeros of $$f$$, respectively. In $$\mathfrak{S}(G)$$ we can define an addition $$(S,f) + (T,g) = (S \cup T,f + g)$$, where $$(f + g)(z) = f(z) + g(z)$$ for $$z \in G \setminus (S \cup T)$$, similarly a multiplication $$(S,f) \cdot (T,g)$$. To simplify notation, we usually write $$f$$ instead of $$(S,f)$$.

The set $$\mathfrak{M}(G)$$ of meromorphic functions on $$G$$ is defined as the subset of $$\mathfrak{S}(G)$$ such that $$R(f) = \emptyset$$.

"Removing removable singularities" yields a surjective function $$\rho : \mathfrak{S}(G) \to \mathfrak{M}(G)$$.

The algebraic operations on $$\mathfrak{S}(G)$$ induce algebraic operations on $$\mathfrak{M}(G)$$ by $$f + g = \rho(f + g)$$, $$f \cdot g = \rho(f \cdot g)$$. It is easy to see that $$\mathfrak{M}(G)$$ with these operations is a field. The multiplicative inverse of $$f \ne 0$$ is $$\rho(\frac{1}{f})$$. In fact, $$f$$ is defined on $$G \setminus P(f)$$ and $$\frac{1}{f}$$ is defined on $$G \setminus (P(f) \cup Z(f))$$. The zeros of $$f$$ are poles of $$\frac{1}{f}$$ and the poles of $$f$$ are removable singularities of $$\frac{1}{f}$$ so that $$\rho(\frac{1}{f})$$ is defined on $$G \setminus Z(f)$$ with poles at the points of $$Z(f)$$ and zeros at the points of $$P(f)$$.